A) \[20(\sqrt{3}+1)\,\,m\]
B) \[\frac{20\sqrt{3}}{\sqrt{3}+1}\,\,m\]
C) \[\frac{40\sqrt{3}}{\sqrt{3}+1}\,\,m\]
D) \[\frac{40\sqrt{3}}{\sqrt{3}-1}\,\,m\]
Correct Answer: D
Solution :
Let PQ (= h) be the height of the tower. Let S and R be the points where the angles subtended are \[45{}^\circ \]and \[60{}^\circ .\] |
In \[\Delta PQR,\] |
\[\tan 60{}^\circ =\frac{PQ}{RQ}\] |
\[\Rightarrow \] \[\sqrt{3}=\frac{h}{x}\] |
\[\Rightarrow \] \[x=\frac{h}{\sqrt{3}}\] ?(i) |
In \[\Delta PQS,\] |
\[\tan 45{}^\circ =\frac{PO}{SQ}=\frac{h}{x+40}\] |
\[\Rightarrow \] \[h=40+x\] |
\[\Rightarrow \] \[h=40+\frac{h}{\sqrt{3}}\] [from Eq. (i)] |
\[h\,\,\left( 1-\frac{1}{\sqrt{3}} \right)=40\] |
\[\Rightarrow \] \[h\,\,\left( \frac{\sqrt{3}-1}{\sqrt{3}}=40 \right)\]\[\Rightarrow \]\[h=\frac{40\sqrt{3}}{\sqrt{3}-1}\,\,m\] |
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